Chemistry 101

The ph of 0.001 of molar solution of benzen carborhylic acid is 3.59. Calculate the dissociation constant of the acid.

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  1. I assume you are talking about benzoic acid.
    Let's call this acid HBz
    HBz ==> H^+ + Bz^-
    Ka = (H^+)(Bz^-)/(HBz)
    pH = 3.59 = -log (H^+).
    Solve for (H^+) and I get approximately 2.6E-4 but you need to do it more accurately than that.
    (H^+) = from above.
    (Bz^-) = (H^+)
    (HBz) = 0.001-(H^+) = ??
    Solve for Ka.

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